OK, I have read two different proofs of the following theorem both of which I can't quite wrap my mind around. So, I tried to write a proof that makes sense to me, and hopefully to others with the same difficulty. Please let me know if this is an accurate proof an how I might fix it. Sorry, about the length I have been thinking about this proof all day! Thanks for your time.

First, we show that $|A|\leq |\mathscr{P}(A)|$ by showing that there exsists a injective function from $A$ to $\mathscr{P}(A)$. Let, $f:A\to \mathscr{P}(A)$ be defined by the function $a\mapsto \{a\} $ that is that every element $a \in A$ is mapped to the singleton set in $\mathscr{P}(A)$ containing the element $a$. Thus we have found an injection from $A$ to $\mathscr{P}(A)$ and therefore $|A|\leq |\mathscr{P}(A)|$.

Now, in order to show that $|A|<|\mathscr{P}(A)|$ we must show that there does not exist a surjective function from $A$ to $\mathscr{P}(A)$. Let, us suppose that there does exist a surjective function $f:A\to \mathscr{P}(A)$ then it is implied (by the nature of this mapping) that $(\forall a\in A)(f(a) \subseteq A)$. Now, let us define a set $B:=\{x\in A : x\notin f(x)\}$. We want to show two things: first we want to show that $B\subseteq A$, and secondly we want that $B \not\subseteq\mathscr{P}(A)$. If the set $B=\emptyset$ then $B\subseteq A$ since the empty set is a subset of all sets. If $B\not =\emptyset$ then $B\subseteq A$ by the definition of $B$. Consequently, there must be an element $x'\in A$ such that $f(x')=B$ or otherwise $B\not\subseteq \mathscr{P}(A)$. In other words there must exist an an element $x'$ in $A$ such that the image of $x'$ under $f$, $f(x')$ , belongs to $B$ in order for $B \subseteq \mathscr{P}(A)$, or otherwise the opposite is true. Since we have already determined that $B \subseteq A$ then it must be true that $x'\in B$ or that $x \notin B$. If $x'\in B$ then $x' \notin f(x')$ which means that $x' \notin B$ since $f(x')=B$. Likewise, if $x' \notin B$ then $x' \in f(x) = B$. So, either way we arrive at a contradiction, and therefore it must be true that $B \not\subseteq \mathscr{P}(A)$. Hence, $f$ is not surjetive.

In conclusion, we showed that $|A|\leq |\mathscr{P}(A)|$, and since the function $f$ was arbitrary mapping it must be true that there does not exist a surjective mapping from $A$ to $\mathscr{P}(A)$. And therefore we have that $|A|< |\mathscr{P}(A)|$ is always true.

I am not sure, if there are any flaws in my logic please let me know. I would like to edit the post so that a detailed proof will be available for other who have struggled with this proof.

3 Answers
3

The argument is basically just a slightly more detailed than usual version of the standard proof, but you’ve made one consistent error throughout, writing $\nsubseteq\wp(A)$ when you actually mean $\notin\operatorname{ran}f$.

We want to show two things: first we want to show that $B\subseteq A$, and secondly we want that $B\nsubseteq\wp(A)$.

No, we want to show that $B\subseteq A$ and that $B\notin\operatorname{ran}f$, thereby showing that the function $f$ is not a surjection. There’s no need to split the proof that $B\subseteq A$ into two cases: you defined $B$ to be $\{x\in A:x\notin f(x)\}$, which automatically makes $B$ a subset of $A$.

Consequently, there must be an element $x'\in A$ such that $f(x')=B$ or otherwise $B\nsubseteq\wp(A)$.

The last bit should be $B\notin\operatorname{ran}f$. We know that $B\subseteq A$ and hence that $B\in\wp(A)$, and we’re assuming that $f$ is a surjection, so that everything in $\wp(A)$ is in the range of $f$. And since $B$ is not in general a collection of subsets of $A$, it cannot in general be a subset of $\wp(A)$; it’s an element of $\wp(A)$. (There are special circumstances in which $B$ might turn out, more or less by accident, to be a subset of $A$).

In other words there must exist an an element $x'$ in $A$ such that the image of $x'$ under $f$, $f(x')$, belongs to $B$ in order for $B\subseteq\wp(A)$, or otherwise the opposite is true.

Again, we know that $B\subseteq A$, i.e., that $B\in\wp(A)$; $B$ is certainly not in general a subset of $\wp(A)$. What you should be saying here is:

In other words, there must be an element $x'$ of $A$ such that the image of $x'$ under $f$, $f(x')$, is $B$ in order for $B\in\operatorname{ran}f$ to be true; otherwise, the opposite is true, and $f$ is not a surjection.

The next sentence has an unnecessary bit whose presence might actually confuse some, if they thought that it actually was relevant:

Since we have already determined that $B\subseteq A$ then it must be true that $x'\in B$ or that $x\notin B$.

The hypothesis is unnecessary: no matter what $x'$ and $B$ are, exactly one of the statements $x'in B$ and $x'\notin B$ is true.

So, either way we arrive at a contradiction, and therefore it must be true that $B\nsubseteq\wp(A)$.

This is the same mistake that you’ve made in several other places: it should read ‘and therefore it must be true that $B\notin\operatorname{ran}f$’.

$\begingroup$@BrianM.Scott Thanks for this! I rewrote the proof below, and it finally make sense. Could you suggest an accessible, yet thorough, Set Theory book that would go well with a reading in Elementary Analysis, and Elementary Abstract Algebra?$\endgroup$
– JimmyJacksonJul 6 '13 at 8:10

One mistake that wasn't covered by Brian's extensive answer, is the formulation of the theorem. It is a common mistake, especially by beginners.

Cantor's theorem tells us there is no surjection from a set onto its power set. In particular there is no bijection.

This is a delicate issue, but under some (advanced) set theoretical assumptions it is possible to have an injection from $A$ to $B$, a surjection from $A$ onto $B$, and yet no bijection between the two sets. But even in that framework Cantor's theorem is provable and true.

$\begingroup$Thanks for this extra information as well, My Real Analysis assumes that I know a lot of things that I don't really know. Do you know any accessible, but rigorous books on set theory that might be useful before diving into Real Analysis. Or should I just keep learning as I go asking questions on here?$\endgroup$
– JimmyJacksonJul 6 '13 at 7:17

First, we show that $|A|\leq |\mathscr{P}(A)|$ by showing that there exsists a injective function from $A$ to $\mathscr{P}(A)$. Let, $f:A\to \mathscr{P}(A)$ be defined by the function $a\mapsto \{a\} $ that is that every element $a \in A$ is mapped to the singleton set in $\mathscr{P}(A)$ containing the element $a$. Thus we have found an injections from $A$ to $\mathscr{P}(A)$ and therefore $|A|\leq |\mathscr{P}(A)|$.

Now in order to show that $|A|<|\mathscr{P}(A)|$ we must show that there does not exist a surjective function from $A$ to $\mathscr{P}(A)$. Let, us suppose that there does exist a surjective function $f:A\to \mathscr{P}(A)$. Now, let us define a set $B:=\{x\in A : x\notin f(x)\}$. We want to show two things: first we want to show that $B\subseteq A$, and secondly we want that $B \notin \text{ran} f$. We know that $ B \subseteq A$ by the definition of $B$. Consequently, there must be an element $x'\in A$ such that $f(x')=B$ or otherwise $B\notin \text{ran} f$. In other words there must exist an an element $x'$ in $A$ such that the image of $x'$ under $f$, $f(x')$ , is equal to $B$ in order for $B \in \text{ran} f$ to be true; otherwise, the opposite is true, $B\notin \text{ran} f$. Now, for any element in $A$ it must be true that $x'\in B$ or that $x' \notin B$; if $x'\in B$ then $x' \notin f(x')$ which means that $x' \notin B$ since $f(x')=B$. Likewise, if $x' \notin B$ then $x' \in f(x) = B$. So, either way we arrive at a contradiction, and therefore it must be true that $B \notin \text{ran} f$. Hence, $f$ is not surjetive.

In conclusion, we showed that $|A|\leq |\mathscr{P}(A)|$, and since the function $f$ was arbitrary mapping it must be true that there does not exist a surjective mapping from $A$ to $\mathscr{P}(A)$. And therefore we have that $|A|< |\mathscr{P}(A)|$ is always true.